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Kirill Smelkov
cpython
Commits
91149821
Commit
91149821
authored
Jan 31, 2003
by
Tim Peters
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Linear-time implementations of {encode,decode}_long.
parent
1a17704f
Changes
1
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Lib/pickle.py
Lib/pickle.py
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Lib/pickle.py
View file @
91149821
...
...
@@ -1285,10 +1285,12 @@ class Unpickler:
class
_EmptyClass
:
pass
# Encode/decode longs.
# Encode/decode longs in linear time.
import
binascii
as
_binascii
def
encode_long
(
x
):
r"""Encode a long to a two's complement little-endi
ng
binary string.
r"""Encode a long to a two's complement little-endi
an
binary string.
>>> encode_long(255L)
'\xff\x00'
>>> encode_long(32767L)
...
...
@@ -1303,14 +1305,46 @@ def encode_long(x):
'\x7f'
>>>
"""
# XXX This is still a quadratic algorithm.
# Should use hex() to get started.
digits
=
[]
while
not
-
128
<=
x
<
128
:
digits
.
append
(
x
&
0xff
)
x
>>=
8
digits
.
append
(
x
&
0xff
)
return
""
.
join
(
map
(
chr
,
digits
))
if
x
==
0
:
return
'
\
x00
'
if
x
>
0
:
ashex
=
hex
(
x
)
assert
ashex
.
startswith
(
"0x"
)
njunkchars
=
2
+
ashex
.
endswith
(
'L'
)
nibbles
=
len
(
ashex
)
-
njunkchars
if
nibbles
&
1
:
# need an even # of nibbles for unhexlify
ashex
=
"0x0"
+
ashex
[
2
:]
elif
ashex
[
2
]
>=
'8'
:
# "looks negative", so need a byte of sign bits
ashex
=
"0x00"
+
ashex
[
2
:]
else
:
# Build the 256's-complement: (1L << nbytes) + x. The trick is
# to find the number of bytes in linear time (although that should
# really be a constant-time task).
ashex
=
hex
(
-
x
)
assert
ashex
.
startswith
(
"0x"
)
njunkchars
=
2
+
ashex
.
endswith
(
'L'
)
nibbles
=
len
(
ashex
)
-
njunkchars
if
nibbles
&
1
:
# need an even # of nibbles for unhexlify
nibbles
+=
1
nbytes
=
nibbles
>>
1
x
+=
1L
<<
(
nbytes
*
8
)
assert
x
>
0
ashex
=
hex
(
x
)
if
x
>>
(
nbytes
*
8
-
1
)
==
0
:
# "looks positive", so need a byte of sign bits
ashex
=
"0xff"
+
x
[
2
:]
if
ashex
.
endswith
(
'L'
):
ashex
=
ashex
[
2
:
-
1
]
else
:
ashex
=
ashex
[
2
:]
assert
len
(
ashex
)
&
1
==
0
binary
=
_binascii
.
unhexlify
(
ashex
)
return
binary
[::
-
1
]
def
decode_long
(
data
):
r"""Decode a long from a two's complement little-endian binary string.
...
...
@@ -1327,15 +1361,12 @@ def decode_long(data):
>>> decode_long("\x7f")
127L
"""
# XXX This is quadratic too.
x
=
0L
i
=
0L
for
c
in
data
:
x
|=
long
(
ord
(
c
))
<<
i
i
+=
8L
if
data
and
ord
(
c
)
>=
0x80
:
x
-=
1L
<<
i
return
x
ashex
=
_binascii
.
hexlify
(
data
[::
-
1
])
n
=
long
(
ashex
,
16
)
if
data
[
-
1
]
>=
'
\
x80
'
:
n
-=
1L
<<
(
len
(
data
)
*
8
)
return
n
# Shorthands
...
...
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